Chapter 12, Problem R12.6RE

(a)

To determine

To explain why the linear model is not appropriate for describing the relationship between price and weight of diamonds.

Answer to Problem R12.6RE

It will not be appropriate.

Explanation of Solution

The scatterplot showing the relationship between the weight and price of round circle internally flawless diamonds with excellent cuts is given in the question. A liner model would not be appropriate for describing the relationship between price and weight of diamonds because the given scatterplot of the price versus weight contains a lot of curvature and thus there appears to be curved relationship between the weight and the price.

(b)

To determine

To explain would an exponential model or a power model describe the relationship better.

Answer to Problem R12.6RE

A power model describe the relationship better

Explanation of Solution

The scatterplot showing the relationship between the weight and price of round circle internally flawless diamonds with excellent cuts is given in the question. The two transformation are given in the question. So, we have,

Transformation $2$ : Exponential model,

In this, we note that the given scatterplot of ln(price) versus weight contains strong curvature and thus it will be not appropriate to use a linear model between the two variables of the scatterplot. Thus it is not appropriate to use a linear model between the weight and ln(price). So, the general linear model to predict ln(price) from weight is as:

  $\mathrm{l}\widehat{n}(\text{Price)}=a+b(\text{Weight})$

Now, taking the exponential on each side, we have,

  $\begin{array}{l}\text{<mover accent="true"><mo>P</mo><mo>^</mo></mover> rice}=e^{\mathrm{l}\widehat{n}(\text{Price)}}\\ =e^{a+b(\text{Weight})}\\ =e^a{e}^{b(\text{Weight})}\end{array}$

Transformation $1$ : Power model,

In this we note that the given scatterplot of ln(weight) and ln(price) does not contain any strong curvature and thus it will be not appropriate to use a linear model between the two variables of the scatterplot. Thus, it will be appropriate to use a linear model between ln(weight) and ln(price). So, the general linear model to predict this is:

  $\mathrm{l}\widehat{n}(\text{Price)}=a+b\ln (\text{Weight})$

Taking the exponential on both the sides as:

  $\begin{array}{l}\text{<mover accent="true"><mo>P</mo><mo>^</mo></mover> rice}=e^{\mathrm{l}\widehat{n}(\text{Price)}}\\ =e^{a+b\ln (\text{Weight})}\\ =e^a{e}^{b\ln (\text{Weight})}\end{array}$

Thus, we conclude that a power model describe the relationship better.

(c)

To determine

To find out which prediction do you think would be better.

Answer to Problem R12.6RE

Power model would be better.

Explanation of Solution

The scatterplot showing the relationship between the weight and price of round circle internally flawless diamonds with excellent cuts is given in the question. The two transformation are given in the question. Thus, we have,

Transformation $2$ : Exponential model,

The general equation of the least square regression line is:

  $\widehat{y}=b_0+b_{1}x$

Thus, form the computer output, we have that the estimate of the constant is given in the row “Constant” and in the column “Coef” as:

  $b_0=8.2709$

The slope $b_1$ is given in the row “Weight” and in the column “Coef” of the given computer output as:

  $b_1=1.3791$

Now replacing the values in the equation we have,

  $\begin{array}{l}\widehat{y}=b_0+b_{1}x\\ =8.2709+1.3791x\end{array}$

Now, take the logarithm in the equation and solve it as:

  $\ln \widehat{y}=8.2709+1.3791x$

Replace xby $2$ ,

  $\begin{array}{l}\ln \widehat{y}=8.2709+1.3791x\\ =8.2709+1.3791(2)\\ =11.0291\end{array}$

Now taking exponential on both sides we have,

  $\begin{array}{l}\widehat{y}=e^{\ln \widehat{y}}\\ =e^{11.0291}\\ =61642.1\end{array}$

The predicted price is $\$61642.1$ .

Transformation $1$ : Power model,

The general equation of the least square regression line is:

  $\widehat{y}=b_0+b_{1}x$

Thus, form the computer output, we have that the estimate of the constant is given in the row “Constant” and in the column “Coef” as:

  $b_0=9.7062$

The slope $b_1$ is given in the row “lnWeight” and in the column “Coef” of the given computer output as:

  $b_1=2.2913$

Now replacing the values in the equation we have,

  $\begin{array}{l}\widehat{y}=b_0+b_{1}x\\ =9.7062+2.2913x\end{array}$

Now, take the logarithm in the equation and solve it as:

  $\ln \widehat{y}=9.7062+2.2913x$

Replace xby $2$ ,

  $\begin{array}{l}\ln \widehat{y}=9.7062+2.2913x\\ =9.7062+2.2913(2)\\ =11.2944\end{array}$

Now taking exponential on both sides we have,

  $\begin{array}{l}\widehat{y}=e^{\ln \widehat{y}}\\ =e^{11.2944}\\ =80370.30\end{array}$

The predicted price is $\$80370.30$ .

Thus, prediction using the power model would be better form the following above results.